# Exponential and Logarithmic problems

• Jul 24th 2007, 10:23 AM
JonathanEyoon
Exponential and Logarithmic problems
I can do such problems like

6^x+1 = 4^2x-1

but I can't figure out how to do

.05(1.15)^X = 5

Help =(
• Jul 24th 2007, 10:29 AM
galactus
$\displaystyle \frac{1}{20}(\frac{23}{20})^{x}=5$

$\displaystyle (\frac{23}{20})^{x}=100$

$\displaystyle xln(\frac{23}{20})=ln(100)$

$\displaystyle x=\frac{2ln(10)}{ln(\frac{23}{20})}$

$\displaystyle \approx{32.95}$
• Jul 24th 2007, 10:32 AM
JonathanEyoon
Quote:

Originally Posted by galactus
$\displaystyle \frac{1}{20}(\frac{23}{20})^{x}=5$

$\displaystyle (\frac{23}{20})^{x}=100$

$\displaystyle xln(\frac{23}{20})=ln(100)$

$\displaystyle x=\frac{2ln(10)}{ln(\frac{23}{20})}$

$\displaystyle \approx{32.95}$

Where did the 100 on the other side come from?
• Jul 24th 2007, 12:15 PM
Krizalid
It's really simple

\displaystyle \begin{aligned} \frac{1} {{20}}\left( {\frac{{23}} {{20}}} \right)^x &= 5\\ 20 \cdot \frac{1} {{20}}\left( {\frac{{23}} {{20}}} \right)^x &= 20 \cdot 5\\ \left( {\frac{{23}} {{20}}} \right)^x &= 100 \end{aligned}
• Jul 24th 2007, 12:20 PM
Krizalid
Quote:

Originally Posted by JonathanEyoon
6^x+1 = 4^2x-1

I assume you forgot the brackets

\displaystyle \begin{aligned} 6^{x + 1} &= 4^{2x - 1}\\ (x + 1)\log 6 &= 2(2x - 1)\log 2\\ x\log 6 + \log 6 &= 4x\log 2 - 2\log 2\\ 4x\log 2 - x\log 6 &= 2\log 2 + \log 6\\ x(4\log 2 - \log 6) &= 2\log 2 + \log 6\\ x &= \frac{{2\log 2 + \log 6}} {{4\log 2 - \log 6}} \end{aligned}